Answer
$$(\sin x)^3 \cos x. $$
Work Step by Step
By making use of Theorem 1 (FTC II), we have $$ G'(x)=\frac{d}{d x} \int_{-2}^{\sin x} t^3 dt =\left(\frac{d}{d u} \int_{-2}^{u} t^3 dt\right)\frac{du}{dx}=(\sin x)^3 (\cos x)\\ =(\sin x)^3 \cos x. $$
